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Proof step 16 from proof violations-swap (previous step)

less#(a, x0) + violations(x0) = m + 1, n = m1 + 1, x0[n] < x0[m1], n < # x0, ∀ m0, n0. violations(x0) = m0 + 1 ∧ x0[n0 + 1] < x0[n0] ∧ n0 + 1 < # x0 → violations(x0[n0, x0[n0 + 1]][n0 + 1, x0[n0]]) = m0 ⊦ n = 0, violations((a ' + x0)[n, x0[n]][n + 1, x0[m1]]) = m, x0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
less#(a, x0) + violations(x0) = m + 1, m1 + 1 < # x0, x0[m1 + 1] < x0[m1], ∀ m0, n0. violations(x0) = m0 + 1 ∧ x0[n0 + 1] < x0[n0] ∧ n0 + 1 < # x0 → violations(x0[n0, x0[n0 + 1]][n0 + 1, x0[n0]]) = m0 ⊦ violations((a ' + x0[m1, x0[m1 + 1]])[2 + m1, x0[m1]]) = m, x0 = []

Rule argument

Additional information

Used simplifier rules: ⊦ m + 1 ≠ 0; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ a + x = a ' + x; (s)% ⊦ (b ' + x)[n + 1, a] = b + x[n, a]; (s)